+4 The vertices of triangle $ABC$ lie on the sides of equilateral triangle $DEF$, as shown. If $CD = 5$, $CE = BD = 2$, and $\angle ACB = 90^\circ$, then find $AE$.
+1768 Since ∠ACB=90∘, we have ∠ACE=30∘. Also, since △DEF is equilateral, ∠DEF=60∘. Then ∠AED=60∘−30∘=30∘. Furthermore, △AED is isosceles since AD=AE, so ∠ADE=30∘. Since the sum of the angles in a triangle is 180∘, we have \begin{align*} \angle AED + \angle ADE + \angle EAD &= 180^\circ \ 30^\circ + 30^\circ + \angle EAD &= 180^\circ \ \angle EAD &= 120^\circ. \end{align*}By the Law of Sines in △ADE, we have \begin{align*} \frac{AD}{\sin \angle EAD} &= \frac{AE}{\sin \angle ADE} \ \frac{AE}{5} &= \frac{AE}{\sin 30^\circ} \ 5 &= \sin 30^\circ \ AE &= \frac{5}{\sin 30^\circ} \ AE &= \boxed{\frac{10}{\sqrt{3}}}. \end{align*}
So the answer is AE = 10/sqrt(3).
